\(\int \frac {1}{x (1+b x)} \, dx\) [279]

Optimal result

Integrand size = 11, antiderivative size = 11 \[ \int \frac {1}{x (1+b x)} \, dx=\log (x)-\log (1+b x) \]

[Out]

ln(x)-ln(b*x+1)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {36, 29, 31} \[ \int \frac {1}{x (1+b x)} \, dx=\log (x)-\log (b x+1) \]

[In]

Int[1/(x*(1 + b*x)),x]

[Out]

Log[x] - Log[1 + b*x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rubi steps \begin{align*} \text {integral}& = -\left (b \int \frac {1}{1+b x} \, dx\right )+\int \frac {1}{x} \, dx \\ & = \log (x)-\log (1+b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (1+b x)} \, dx=\log (x)-\log (1+b x) \]

[In]

Integrate[1/(x*(1 + b*x)),x]

[Out]

Log[x] - Log[1 + b*x]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09

method	result	size
default	\(\ln \left (x \right )-\ln \left (b x +1\right )\)	\(12\)
norman	\(\ln \left (x \right )-\ln \left (b x +1\right )\)	\(12\)
parallelrisch	\(\ln \left (x \right )-\ln \left (b x +1\right )\)	\(12\)
meijerg	\(\ln \left (x \right )+\ln \left (b \right )-\ln \left (b x +1\right )\)	\(14\)
risch	\(\ln \left (-x \right )-\ln \left (b x +1\right )\)	\(14\)

[In]

int(1/x/(b*x+1),x,method=_RETURNVERBOSE)

[Out]

ln(x)-ln(b*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (1+b x)} \, dx=-\log \left (b x + 1\right ) + \log \left (x\right ) \]

[In]

integrate(1/x/(b*x+1),x, algorithm="fricas")

[Out]

-log(b*x + 1) + log(x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x (1+b x)} \, dx=\log {\left (x \right )} - \log {\left (x + \frac {1}{b} \right )} \]

[In]

integrate(1/x/(b*x+1),x)

[Out]

log(x) - log(x + 1/b)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (1+b x)} \, dx=-\log \left (b x + 1\right ) + \log \left (x\right ) \]

[In]

integrate(1/x/(b*x+1),x, algorithm="maxima")

[Out]

-log(b*x + 1) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x (1+b x)} \, dx=-\log \left ({\left | b x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/x/(b*x+1),x, algorithm="giac")

[Out]

-log(abs(b*x + 1)) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x (1+b x)} \, dx=-2\,\mathrm {atanh}\left (2\,b\,x+1\right ) \]

[In]

int(1/(x*(b*x + 1)),x)

[Out]

-2*atanh(2*b*x + 1)